<u>Answer:</u> The final volume of the oxygen gas is 4.04 L
<u>Explanation:</u>
To calculate the final temperature of the system, we use the equation given by Charles' Law. This law states that volume of the gas is directly proportional to the temperature of the gas at constant pressure and number of moles.
Mathematically,
where,
are the initial volume and temperature of the gas.
are the final volume and temperature of the gas.
We are given:
Putting values in above equation, we get:
Hence, the final volume of the oxygen gas is 4.04 L
Benedict's solution is used to test simple sugars, such as glucose. It is blue solution, when sugar is present, it turns to orange / brick red. Depends on the concentration of sugar.
Answer:
412.1kJ
Explanation:
For the reaction , from the question -
4Fe (s) + 3O₂ (g) → 2Fe₂O₃ (s)
Δ Hrxn = Δ H°f (products) - Δ H°f (reactants)
In case the compound is in its standard state , enthalphy of formation is zero
Hence ,
for the above reaction ,
ΔHrxn =( 2 * Δ H° (Fe₂O₃ )) - [ ( 4 *Δ H° Fe ) + (3 * Δ H° O₂ )]
The value for Δ H°(Fe₂O₃ ) = - 824.2kJ/mol
Δ H° Fe = 0
Δ H° O₂ = 0
Putting in the above equation ,
ΔH rxn = ( 2 * Δ H° (Fe₂O₃ )) - 0
ΔHrxn = 2× - 824.2 kJ / mol = - 1648.4 kJ/mol
- 1648.4 kJ/mol , this much heat is released by the buring of 4 mol of Fe.
Hence ,
for 1 mol of Fe ,
- 1648.4 kJ/mol / 4 = 412.1kJ
(D) It shares to electrons with another non metal. Take H2O for example. Oxygen has 6 valence electrons, Hydrogen has one. Each valence electron on hydrogen will share its electron with one of the two remaining unpaired electrons of Oxygen.
<u>Answer:</u> The activation energy for the reaction is 40.143 kJ/mol
<u>Explanation:</u>
To calculate activation energy of the reaction, we use Arrhenius equation for two different temperatures, which is:
![\ln(\frac{K_{317K}}{K_{278K}})=\frac{E_a}{R}[\frac{1}{T_1}-\frac{1}{T_2}]](https://tex.z-dn.net/?f=%5Cln%28%5Cfrac%7BK_%7B317K%7D%7D%7BK_%7B278K%7D%7D%29%3D%5Cfrac%7BE_a%7D%7BR%7D%5B%5Cfrac%7B1%7D%7BT_1%7D-%5Cfrac%7B1%7D%7BT_2%7D%5D)
where,
= equilibrium constant at 317 K = 
= equilibrium constant at 278 K = 
= Activation energy = ?
R = Gas constant = 8.314 J/mol K
= initial temperature = 278 K
= final temperature = 317 K
Putting values in above equation, we get:
![\ln(\frac{3.050\times 10^8}{3.600\times 10^{7}})=\frac{E_a}{8.314J/mol.K}[\frac{1}{278}-\frac{1}{317}]\\\\E_a=40143.3J/mol=40.143kJ/mol](https://tex.z-dn.net/?f=%5Cln%28%5Cfrac%7B3.050%5Ctimes%2010%5E8%7D%7B3.600%5Ctimes%2010%5E%7B7%7D%7D%29%3D%5Cfrac%7BE_a%7D%7B8.314J%2Fmol.K%7D%5B%5Cfrac%7B1%7D%7B278%7D-%5Cfrac%7B1%7D%7B317%7D%5D%5C%5C%5C%5CE_a%3D40143.3J%2Fmol%3D40.143kJ%2Fmol)
Hence, the activation energy for the reaction is 40.143 kJ/mol
